A-level · Mechanics · Beta

Forces and Newton’s laws

Connect a force diagram to resultant force, acceleration and velocity.

Change the forces and mass, then trace the motion in time. Right is positive.

Fnet=12−4=8 NF_{\rm net}=12-4=8\,\mathrm{N}
Forces on a particle on a smooth horizontal surfaceFree-body diagram: forces on the particle4 kg12 N4 NR = 39.2 Nmg = 39.2 NRight is positive. Zero-force arrows are omitted.

Scroll the diagram sideways to see every label.

The acceleration points in the direction of the resultant force.

Resultant force (N)
8
Acceleration (m/s²)
2
Velocity at 1 s (m/s)
2
Displacement (m)
1
Vertical resultant (N)
0
Build the equations

R−mg=0,12−4=4aR-mg=0,\quad 12-4=4a

a=84,v=u+at,s=ut+12at2a=\frac{8}{4},\quad v=u+at,\quad s=ut+\tfrac12at^2

Smooth horizontal surface; constant horizontal forces; no air resistance. Gravity g = 9.8 m/s². Reaction and weight balance vertically but are not a Newton’s third-law pair: both act on this particle. The free-body diagram is not a position animation. Values rounded to 3 decimal places.

Predict, test and explain

Step 1 of 3

Predict

If the resultant force is zero, must the particle be at rest?

InvestigateBalance the horizontal forces. Must the particle be at rest?

Explore and compare

Choose the investigation above. Predict what will change before you move a control.

Try it in the simulation. Change one thing at a time.

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Hints, self-check and connections
Try a self-check

Fixed values, separate from the diagram controls.

If the resultant force is zero, must the particle be at rest?

12−4=4a12-4=4a

Hint

Use the resultant force, not just one applied force, in F = ma.

Reveal answer

Zero resultant gives zero acceleration, not necessarily zero velocity.

a=2 m s−2a=2\,\mathrm{m\,s^{-2}}

Build an explanation

Hint 1

Use the resultant force, not just one applied force, in F = ma.

Hint 2

Zero resultant gives zero acceleration, not necessarily zero velocity.

Hint 3

State the idealised assumptions before applying the equations.

Worked example

This example uses fixed values, separate from the diagram controls.

A 4 kg particle has forces of 12 N right and 4 N left. Find its acceleration.

  1. Take right as positive. The resultant is 12 − 4 = 8 N.
  2. Apply F = ma: 8 = 4a.
  3. Divide by 4 to obtain a = 2 m/s² to the right.

Answer: 2 m/s² to the right.

Connect this idea

Watch out: Zero resultant gives zero acceleration, not necessarily zero velocity.

Specification and learning route

AQA R1, R2, R3 · Edexcel Mechanics 8.1, Mechanics 8.2, Mechanics 8.3

AS and A-level. This model illustrates selected content; references are not a claim of full coverage or exam-board endorsement.

Useful starting knowledge: Signed numbers; Force and mass units; Constant acceleration.

AQA specification · Edexcel specification